Problem: $f(x,y) = \dfrac{\cos(y)}{x}$ What is $\dfrac{\partial f}{\partial y}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{\sin(y)}{x^2}$ (Choice B) B $\dfrac{\cos(y)}{x^2}$ (Choice C) C $\dfrac{-\sin(y)}{x}$ (Choice D) D $\dfrac{-\cos(y)}{x^2}$
Solution: We want to find $\dfrac{\partial f}{\partial y}$, which is the partial derivative of $f$ with respect to $y$. When we take a partial derivative with respect to $y$, we treat $x$ as if it were a constant. Here, $f(x, y)$ only has one term. $\begin{aligned} &\dfrac{\partial}{\partial y} \left[ \dfrac{\cos(y)}{x} \right] = \dfrac{1}{x} \dfrac{\partial}{\partial y} \left[ \cos(y)\right] = \dfrac{-\sin(y)}{x} \end{aligned}$ In conclusion: $\dfrac{\partial f}{\partial y} = \dfrac{-\sin(y)}{x}$